[3 10 5 , -2 -3 -4, 3 5 7] find the eigen vectors of the matrix
Answers
Answer:
Step-by-step explanation:
Calculating the eigenvalues would be the first step, in my opinion. For that, I require the characteristic equation. It is obvious that |AI|=0 needs to be resolved. I will, however, adopt the form.
λ3−aλ2+bλ−c=0
And for a quicker outcome, consider the coefficients. Here, the trace, the sum of minors along the diagonal, and the determinant of A are used to determine what a, b, and c are. As a result, a=7, b=16, and c=12.
The indicative formula will be,
λ3–7λ2+16λ−12=0
(1)
The following outcome can be obtained using the rational root theorem or mere guesswork (for the lazy physicists):
(λ−3)(λ−2)2=0
So we have a degenerate eigenvalue here. A matrix with distinct eigenvalues can always be diagonalized because of a simple idea that with each eigenvalue is relate an independent basis eigenvector. And these basis vectors spans the complete eigenspace. The problem with degenerates is that they might have independent basis less than the dimension of the eigenspace. The algebraic multiplicity here for eigenvalue 3 which is distinct is 1 and that for eigenvalue 2 , which is degenerate, is 2 .
The key concept here is that for a matrix to be diagonalizable, an eigenvalue's algebraic multiplicity must match the number of independent bases that make up that eigenvalue, or the algebraic multiplicity must match the geometric multiplicity. This means that they can be regarded as equivalent if a fictitious diagonalized matrix exists with eigenvectors that cover the whole eigenspace of the supplied matrix A and perform the same transformation.
However, in this instance, we obtain a row decreased Echelon when we take into account the degenerate =2.
The rank 2 of this matrix is 2. By using x3=t as a parameter, it is possible to find the basis vectors, which means that the geometric multiplicity is 1 for =2.
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Answer:
Step-by-step explanation:
Calculating the eigenvalues would be the first step, in my opinion. For that, I require the characteristic equation. It is obvious that |AI|=0 needs to be resolved. I will, however, adopt the form.
λ3−aλ2+bλ−c=0
And for a quicker outcome, consider the coefficients. Here, the trace, the sum of minors along the diagonal, and the determinant of A are used to determine what a, b, and c are. As a result, a=7, b=16, and c=12.
The indicative formula will be,
λ3–7λ2+16λ−12=0
The following outcome can be obtained using the rational root theorem or mere guesswork (for the lazy physicists):
(λ−3)(λ−2)2=0
So we have a degenerate eigenvalue here. A matrix with distinct eigenvalues can always be diagonalized because of a simple idea that with each eigenvalue is relate an independent basis eigenvector. And these basis vectors spans the complete eigenspace. The problem with degenerates is that they might have independent basis less than the dimension of the eigenspace. The algebraic multiplicity here for eigenvalue 3 which is distinct is 1 and that for eigenvalue 2 , which is degenerate, is 2 .
The key concept here is that for a matrix to be diagonalizable, an eigenvalue's algebraic multiplicity must match the number of independent bases that make up that eigenvalue, or the algebraic multiplicity must match the geometric multiplicity. This means that they can be regarded as equivalent if a fictitious diagonalized matrix exists with eigenvectors that cover the whole eigenspace of the supplied matrix A and perform the same transformation.
However, in this instance, we obtain a row decreased Echelon when we take into account the degenerate =2.
The rank 2 of this matrix is 2. By using x3=t as a parameter, it is possible to find the basis vectors, which means that the geometric multiplicity is 1 for =2.
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https://brainly.in/question/18105206
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